hi, the paper by Mei et al., 2007 may answer your question. It can be found at http://www.who.int/bulletin/volumes/85/6/06-034421.pdf

Thank you very much for the link. I have another question: I was just wondring what impact, if any, would the data quality have on the values of standard deviation - wider or narrower standard deviations? Also, if the WHO Multicentre Growth Reference Study followed children, why was it not possible to obtain the mean for the reference - i.e. mean weight-for-height z-score for example? Is it because the values (for example, weight-for-height z-score) is not normally distributed? Also, why does the WHO Growth Reference not have percentage of median? I have been told that the reason for this is that z-score is better than percentage of median because zcore takes into account the standard deviation of the reference population. Is that the only reason?

Standard deviation score and percentage of the reference median I will attempt to explain part of your question by first of all giving definitions of what standard deviation and percentage of the median mean. What is important is to know that each of these methods can be used for comparing a child’s measurement with reference values of a normal healthy child. A standard deviation (SD) also known as a z score is the number of standard deviations below or above the reference median value of the reference population. For example, if a child is has a z score of -3SD, it means that this child has a z score of -3SD below the population reference median. In a normal population, 95% of the reference population has z scores between +2SD and -2SD. So for children who fall above or below this, we term them undernourished. Percentage of the median on the other hand expresses the percentage of the child’s measurement as a percent of the reference population (normal child of the same age). However, we cannot tell how many children in a population would be expected to be above or below each percentage of the median because we don’t know the distribution around the median. Difference between the two Where as both can be used to classify nutritional status, the z score has an advantage over the percentage of the median because z scores are normally distributed (e.g, for weight for age, this means that at each age there is a normal distribution of weights around the median value is known) allowing use of analytical procedures that assume normality (for example t-tests) to allow comparison of the calculated prevalence of low anthropometry with the reference population. This means that a fixed z score corresponds to a fixed height or weight difference for children of a similar age. E.g If lets say Jane is 10 years and weighs 6 kg, and has height y cm to calculate Jane’s SD score of weight for age: =current weight- median of ref population/standard deviation of ref population. Let’s say the answer is -2.1SD classifying Jane as moderately malnourished. On the other hand, percentage of the median as straight forward as you hear it, has limitations. E.g If lets say Jane is 10 years and weighs 6 kg, and has height y cm to calculate her percentage weight for age we just express her current weight over the median reference weight of a healthy child of a similar age*100 i.e (current weight/median of ref population*100). The answer may be 89% classifying Jane as almost normal. So percentage of the median may cause misclassifications. Further more, if you want to compare the calculated prevalence of underweight in a population, you cannot compare it with the reference population because we don’t know the expected proportion of the population who fall below a given percent of the median. Also, for each nutritional status indicator, the cut off using percentage of the median is not the same e.g, low weight-for-height is below 80% while height-for- age is below 90%. All this can be confusing……….so why not use z score? So why don’t the new growth standards use percent of the median? Because we know the distribution around the median and this gives a more accurate answer and also allows population comparison with the reference population. And the z score is pretty straight forward Anyone else please feel free to correct whatever I may not have properly expressed to answer the question. this paper may also be useful: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2486530/

It is easy to get the percentage of median. Find the median (p50) weight for the child's height and sex, divide this by the measured weight, multiply by 100. The use of percentage of median has fallen out of fashion. I think this is because it lacks a clear meaning with regard to the reference population ... -2 z-scores (e.g.) is threshold below which about 2.3% of the reference population lies. There is no fixed proportion associated with 80% W/H.